The ratio of final to initial pressures of a gas when `u_(rms)` of a gas in a container is increased from `5 xx 10^(4) cm sec^(-1)" to " 10 xx 10^(4) cm sec^(-1)`

Explanation

To solve the problem of finding the ratio of final to initial pressures of a gas when the root mean square (RMS) speed of the gas changes, we can follow these steps:

### Step 1: Understand the relationship between RMS speed and pressure
The root mean square speed (\(u_{rms}\)) of a gas is given by the formula:
\[ u_{rms} = \sqrt{\frac{3RT}{M}} \]
where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas.

### Step 2: Establish the initial and final RMS speeds
Given:
- Initial RMS speed, \(u_{rms1} = 5 \times 10^4 \, \text{cm/s}\)
- Final RMS speed, \(u_{rms2} = 10 \times 10^4 \, \text{cm/s}\)

### Step 3: Relate RMS speed to pressure
From the formula for \(u_{rms}\), we can see that \(u_{rms}\) is proportional to the square root of pressure when volume and number of moles are constant:
\[ u_{rms} \propto \sqrt{P} \]
This means:
\[ \frac{u_{rms1}}{u_{rms2}} = \sqrt{\frac{P_1}{P_2}} \]

### Step 4: Substitute the values of RMS speeds
Substituting the values of \(u_{rms1}\) and \(u_{rms2}\):
\[ \frac{5 \times 10^4}{10 \times 10^4} = \sqrt{\frac{P_1}{P_2}} \]
This simplifies to:
\[ \frac{1}{2} = \sqrt{\frac{P_1}{P_2}} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{1}{2}\right)^2 = \frac{P_1}{P_2} \]
\[ \frac{1}{4} = \frac{P_1}{P_2} \]

### Step 6: Find the ratio of final to initial pressures
Rearranging gives:
\[ \frac{P_2}{P_1} = 4 \]
Thus, the ratio of final pressure to initial pressure is:
\[ \frac{P_2}{P_1} = 4:1 \]

### Final Answer
The ratio of final to initial pressures of the gas is \(4:1\).

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